Answer
Characteristic Equation: $\lambda^2-8\lambda+3$
$\lambda = 4 \pm \sqrt{13}$
Work Step by Step
1.) To solve, for the characteristic equation, expand $det(A-\lambda I)$ and simplify
$(5-\lambda)(3-\lambda)-(-3)(-4)$
$=(15-5\lambda-3\lambda+\lambda^2)-(12)$
$=\lambda^2-8\lambda+3$
2.) Set the characteristic equation equal to zero to solve for the eigenvalues
$\lambda^2-8\lambda+3=0$
Note: Use the quadratic equation to solve for the eigenvalues
$\lambda=(-b \pm \sqrt{b^2-4ac})/2a$
$\lambda=(8 \pm \sqrt{(-8)^2-4(1)(3)})/2(1)$
$\lambda=(8 \pm \sqrt{64-12})/2$
$\lambda=(8 \pm \sqrt{52})/2$
$\lambda=(8 \pm 2\sqrt{13})/2$
$\lambda=4 \pm \sqrt{13}$
